iOS6 - 社会框架 - SLComposeViewController如何回退到iOS5的TWTweetComposeViewController？

Question

我有一块一块的Twitter共享代码，对iOS6的正常工作，但我需要的应用程序很好地回落到iOS5的太...

它看起来像这样：

- (void) shareOnTwitter 
{ 
    if([SLComposeViewController instanceMethodForSelector:@selector(isAvailableForServiceType)] != nil) 
    { 
     if ([SLComposeViewController isAvailableForServiceType:SLServiceTypeTwitter]) 
     { 
      NSLog(@"twitter available"); 
      SLComposeViewController *composeViewController = [SLComposeViewController composeViewControllerForServiceType:SLServiceTypeTwitter]; 
      [composeViewController setInitialText:self.sharingText]; 
      [self.sharingController presentViewController:composeViewController animated:YES completion:nil]; 
     } 
     else 
     { 
      NSLog(@"twitter not available!"); 
     } 
    } 
    else 
    { 
     // SLComposeViewController not available, this is most likely <iOS6, what to do here? 
    } 
} 

SO，我如何在iOS5中很好地回退（我假设我需要TWTweetComposeViewController），以便我可以在iOS5中使用本地Twitter？

编辑： 最后，我仍然懒于回退到TWTweetComposeViewController，所以我决定简单地回退：iOS6原生tweet - >已安装的twitter应用程序 - > web网址。 这是我放在一起的功能，希望它可以帮助别人：

+(BOOL)isSocialFrameworkAvailable 
{ 
    // whether the iOS6 Social framework is available? 
    return NSClassFromString(@"SLComposeViewController") != nil; 
} 

- (void) shareOnTwitterWithText:(NSString*)text andURL:(NSString*)url andImageName:(NSString*)imageName 
{ 
    // prepare the message to be shared 
    NSString *combineMessage = [NSString stringWithFormat:@"%@ %@", text, url]; 
    NSString *escapedMessage = [combineMessage stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; 
    NSString *appURL = [NSString stringWithFormat:@"twitter://post?message=%@", escapedMessage]; 

    if([SocialManager isSocialFrameworkAvailable] && [SLComposeViewController isAvailableForServiceType:SLServiceTypeTwitter]) 
    { 
     // user has setup the iOS6 twitter account 

     SLComposeViewController *composeViewController = [SLComposeViewController composeViewControllerForServiceType:SLServiceTypeTwitter]; 
     [composeViewController setInitialText:text]; 
     if([UIImage imageNamed:imageName]) 
     { 
      [composeViewController addImage:[UIImage imageNamed:imageName]]; 
     } 
     if(url) 
     { 
      [composeViewController addURL:[NSURL URLWithString:url]]; 
     } 
     [self.sharingController presentViewController:composeViewController animated:YES completion:nil]; 
    } 
    else 
    { 
     // else, we have to fallback to app or browser 
     if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:appURL]]) 
     { 
      // twitter app available! 
      [[UIApplication sharedApplication] openURL:[NSURL URLWithString:appURL]]; 
     } 
     else 
     { 
      // worse come to worse, open twitter page in browser 
      NSString *web = [NSString stringWithFormat:@"https://twitter.com/intent/tweet?text=%@", escapedMessage]; 
      [[UIApplication sharedApplication] openURL:[NSURL URLWithString:web]]; 
     } 
    } 
} 

Answer 1

您需要薄弱环节的Twitter和社会的框架，并在代码检查框架是否可用。弱链接是这样完成的：

• 点击Xcode中的项目，选择目标，构建阶段，然后链接二进制与图书馆
• 确保Social.framework设置为可选，如果你想运行此在iOS 5.x或更早应用
• 确保Twitter.framework设置为可选，如果你想在iOS 4.x中运行这个或更早

我喜欢创造简单的类的功能，以确定哪些框架可用。这可能是这个样子：

+(BOOL)isTwitterAvailable { 
    return NSClassFromString(@"TWTweetComposeViewController") != nil; 
} 

+(BOOL)isSocialAvailable { 
    return NSClassFromString(@"SLComposeViewController") != nil; 
} 

你的“推特”的代码可能是这个样子：

if ([SomeClass isSocialAvailable]) { 
    // code to tweet with SLComposeViewController 
} else if ([SomeClass isTwitterAvailable]) { 
    // code to tweet with TWTweetComposeViewController 
} else { 
    // Twitter not available, or open a url like https://twitter.com/intent/tweet?text=tweet%20text 
} 

Answer 2

不知道这些运行时操作多么昂贵的，但在这样做，因为没有什么危害此应用程序正在运行时此状态更改的机会为零：

+ (BOOL)isTwitterAvailable 
{ 
    static BOOL available; 
    static dispatch_once_t onceToken; 
    dispatch_once(&onceToken, ^{ 
     available = NSClassFromString(@"TWTweetComposeViewController") != nil;   
    }); 
    return available; 
} 
+ (BOOL)isSocialAvailable 
{ 
    static BOOL available; 
    static dispatch_once_t onceToken; 
    dispatch_once(&onceToken, ^{ 
     available = NSClassFromString(@"SLComposeViewController") != nil;   
    }); 
    return available; 
}